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Given topological spaces $X_1, X_2, \dotsc, X_n, Y$, consider a multivariable function $f : \prod_{i = 1}^nX_i \to Y$ such that for any $(x_1, x_2, \dotsc, x_n) \in \prod_{i = 1}^nX_i$, the functions in the family $\{x \mapsto f(x_1, \dotsc, x_{i - 1}, x, x_{i + 1}, \dotsc, x_n)\}_{i = 1}^n$ are all continuous. Must $f$ itself be continuous? It seems to be true, so is there a Theorem that proves this?

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Seek and ye shall find: math.stackexchange.com/questions/98831/… –  Byron Schmuland Feb 7 '13 at 13:33

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The answer is "Yes" if you give the product $\prod X_i$ the slice (or sometimes called "cross topology").

The answer is "No" if you give the product $\prod X_i$ the usual product topology. In this case, $f$ is "separately continuous" but not necessarily continuous. The standard example is the function $f:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ defined by $f(x,y)=\frac{2xy}{x^2+y^2}$ for $(x,y)\neq (0,0)$ and $f(0,0)=(0,0)$. This function is continuous everywhere except $(0,0)$ but is continuous in each variable.

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Didn't see the previous question...oh well. I suppose the cross topology isn't mentioned there. –  Jeremy Feb 7 '13 at 13:40
    
What do you mean by "continuous in each variable"? Did you mean the same thing as in my question, where $y \mapsto f(x_0, y)$ is continuous for all constant $x_0$ and vice versa? –  Herng Yi Feb 7 '13 at 14:12
    
@HerngYi: Yes; this is called separate continuity, and it is weaker than continuity, which is sometimes called joint continuity. Quite a lot of work has been done on separate versus joint continuity; for more information see this answer and this answer and the references cited therein. –  Brian M. Scott Feb 7 '13 at 22:01
    
Thanks, I have read Piotrowski's paper, it was surprisingly easy to understand! Now I have a better idea of the gap between separate and joint continuity. –  Herng Yi Feb 8 '13 at 0:19

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